introduction to mechanics energy...

Introduction to MechanicsEnergy Conservation

Lana Sheridan

De Anza College

Mar 22, 2018

Last time

• conservative forces and potential energy

• energy diagrams

• mechanical energy

• energy conservation

Overview

• energy conservation

• more practice with energy conservation

Energy Conservation

Energy conservation for a system can be expressed as:

Wext = ∆K + ∆U

The work done by external forces includes work done bynonconservative forces and applied forces.

Isolated System: Energy Conservation

A rock is dropped from rest a height h.

By considering energy, find an expression for how fast it is movingjust before it hits the ground.

System: rock + Earth

Let point i be the moment it is dropped; f be just before it strikesthe ground. Let y = 0 be the ground level.

∆K + ∆U = 0

(Kf −��0

Ki ) + (��>0

Uf − Ui ) = 0

Kf = Ui

1

2mv2 = mgh

v =√

2gh


Quick Quiz 8.41 Three identical balls are thrown from the top ofa building, all with the same initial speed. As shown, the first isthrown horizontally, the second at some angle above thehorizontal, and the third at some angle below the horizontal.Neglecting air resistance, rank the speeds of the balls at theinstant each hits the ground, from largest to smallest.

216 Chapter 8 Conservation of Energy

The left side represents a sum of changes of the energy stored in the system. The right-hand side is zero because there are no transfers of energy across the bound-ary of the system; the book–Earth system is isolated from the environment. We devel-oped this equation for a gravitational system, but it can be shown to be valid for a system with any type of potential energy. Therefore, for an isolated system,

DK 1 DU 5 0 (8.6)

(Check to see that this equation is contained within Eq. 8.2.) We defined in Chapter 7 the sum of the kinetic and potential energies of a sys-tem as its mechanical energy:

Emech ; K 1 U (8.7)

where U represents the total of all types of potential energy. Because the system under consideration is isolated, Equations 8.6 and 8.7 tell us that the mechanical energy of the system is conserved:

DEmech 5 0 (8.8)

Equation 8.8 is a statement of conservation of mechanical energy for an iso-lated system with no nonconservative forces acting. The mechanical energy in such a system is conserved: the sum of the kinetic and potential energies remains constant: Let us now write the changes in energy in Equation 8.6 explicitly:

(Kf 2 Ki) 1 (Uf 2 Ui) 5 0

Kf 1 Uf 5 Ki 1 Ui (8.9)

For the gravitational situation of the falling book, Equation 8.9 can be written as12mvf

2 1 mgyf 5 12mvi

2 1 mgyi

As the book falls to the Earth, the book–Earth system loses potential energy and gains kinetic energy such that the total of the two types of energy always remains constant: Etotal,i 5 Etotal, f . If there are nonconservative forces acting within the system, mechanical energy is transformed to internal energy as discussed in Section 7.7. If nonconservative forces act in an isolated system, the total energy of the system is conserved although the mechanical energy is not. In that case, we can express the conservation of energy of the system as

DEsystem 5 0 (8.10)

where Esystem includes all kinetic, potential, and internal energies. This equation is the most general statement of the energy version of the isolated system model. It is equivalent to Equation 8.2 with all terms on the right-hand side equal to zero.

Q uick Quiz 8.3 A rock of mass m is dropped to the ground from a height h. A second rock, with mass 2m, is dropped from the same height. When the second rock strikes the ground, what is its kinetic energy? (a) twice that of the first rock (b) four times that of the first rock (c) the same as that of the first rock (d) half as much as that of the first rock (e) impossible to determine

Q uick Quiz 8.4 Three identical balls are thrown from the top of a building, all with the same initial speed. As shown in Figure 8.3, the first is thrown hori-zontally, the second at some angle above the horizontal, and the third at some angle below the horizontal. Neglecting air resistance, rank the speeds of the balls at the instant each hits the ground.

Mechanical energy Xof a system

The mechanical energy of Xan isolated system with

no nonconservative forces acting is conserved.

The total energy of an Xisolated system is conserved.

Figure 8.3 (Quick Quiz 8.4) Three identical balls are thrown with the same initial speed from the top of a building.

21

3

Pitfall Prevention 8.2Conditions on Equation 8.6 Equa-tion 8.6 is only true for a system in which conservative forces act. We will see how to handle nonconserva-tive forces in Sections 8.3 and 8.4.

(A) 2, 1, 3

(B) 3, 1, 2

(C) 1, 2, 3

(D) all the same

2Adapted from Serway & Jewett, page 216.


Quick Quiz 8.41 Three identical balls are thrown from the top ofa building, all with the same initial speed. As shown, the first isthrown horizontally, the second at some angle above thehorizontal, and the third at some angle below the horizontal.Neglecting air resistance, rank the speeds of the balls at theinstant each hits the ground, from largest to smallest.


The left side represents a sum of changes of the energy stored in the system. The right-hand side is zero because there are no transfers of energy across the bound-ary of the system; the book–Earth system is isolated from the environment. We devel-oped this equation for a gravitational system, but it can be shown to be valid for a system with any type of potential energy. Therefore, for an isolated system,

DK 1 DU 5 0 (8.6)

(Check to see that this equation is contained within Eq. 8.2.) We defined in Chapter 7 the sum of the kinetic and potential energies of a sys-tem as its mechanical energy:

Emech ; K 1 U (8.7)

where U represents the total of all types of potential energy. Because the system under consideration is isolated, Equations 8.6 and 8.7 tell us that the mechanical energy of the system is conserved:

DEmech 5 0 (8.8)

Equation 8.8 is a statement of conservation of mechanical energy for an iso-lated system with no nonconservative forces acting. The mechanical energy in such a system is conserved: the sum of the kinetic and potential energies remains constant: Let us now write the changes in energy in Equation 8.6 explicitly:

(Kf 2 Ki) 1 (Uf 2 Ui) 5 0

Kf 1 Uf 5 Ki 1 Ui (8.9)

For the gravitational situation of the falling book, Equation 8.9 can be written as12mvf

2 1 mgyf 5 12mvi

2 1 mgyi

As the book falls to the Earth, the book–Earth system loses potential energy and gains kinetic energy such that the total of the two types of energy always remains constant: Etotal,i 5 Etotal, f . If there are nonconservative forces acting within the system, mechanical energy is transformed to internal energy as discussed in Section 7.7. If nonconservative forces act in an isolated system, the total energy of the system is conserved although the mechanical energy is not. In that case, we can express the conservation of energy of the system as

DEsystem 5 0 (8.10)

where Esystem includes all kinetic, potential, and internal energies. This equation is the most general statement of the energy version of the isolated system model. It is equivalent to Equation 8.2 with all terms on the right-hand side equal to zero.

Q uick Quiz 8.3 A rock of mass m is dropped to the ground from a height h. A second rock, with mass 2m, is dropped from the same height. When the second rock strikes the ground, what is its kinetic energy? (a) twice that of the first rock (b) four times that of the first rock (c) the same as that of the first rock (d) half as much as that of the first rock (e) impossible to determine

Q uick Quiz 8.4 Three identical balls are thrown from the top of a building, all with the same initial speed. As shown in Figure 8.3, the first is thrown hori-zontally, the second at some angle above the horizontal, and the third at some angle below the horizontal. Neglecting air resistance, rank the speeds of the balls at the instant each hits the ground.

Mechanical energy Xof a system

The mechanical energy of Xan isolated system with

no nonconservative forces acting is conserved.

The total energy of an Xisolated system is conserved.

Figure 8.3 (Quick Quiz 8.4) Three identical balls are thrown with the same initial speed from the top of a building.

21

3

Pitfall Prevention 8.2Conditions on Equation 8.6 Equa-tion 8.6 is only true for a system in which conservative forces act. We will see how to handle nonconserva-tive forces in Sections 8.3 and 8.4.

(A) 2, 1, 3

(B) 3, 1, 2

(C) 1, 2, 3

(D) all the same ←

2Adapted from Serway & Jewett, page 216.

Energy Conservation and Energy Diagrams

Image a ball on a track, released from rest at point A. Suppose wecan ignore friction and air resistance.

8–5 POTENTIAL ENERGY CURVES AND EQUIPOTENTIALS 227

ACTIVE EXAMPLE 8–3 Marathon Man: Find the Height of the HillAn 80.0-kg jogger starts from rest and runs uphill into a stiff breeze. At the top of thehill the jogger has done the work air resistance has done thework and the jogger’s speed is 3.50 m/s. Find the height of the hill.

Solution (Test your understanding by performing the calculations indicated in each step.)

1. Write the initial mechanical energy,

2. Write the final mechanical energy,

3. Set equal to

4. Use to solve for h:

5. Calculate the total nonconservative work:

6. Substitute numerical values to determine h:

InsightAs usual when dealing with energy calculations, our final result is independent ofthe shape of the hill.Your TurnSuppose the jogger’s mass had been 90.0 kg rather than 80.0 kg. What would be theheight of the hill in this case?(Answers to Your Turn problems are given in the back of the book.)

h = 16.7 m

Wnc = Wnc1 + Wnc2 = 13,600 J

h = 1Wnc - 12 mv22/mgWnc = ¢E

Wnc = ¢E = mgh + 12 mv2¢E:Wnc

Ef = Uf + Kf = mgh + 12 mv2Ef:

Ei = Ui + Ki = 0 + 0 = 0Ei:

Wnc2 = -4420 J,Wnc1 = +1.80 * 104 J,

h

v

y

v = 0

i

f

0

▲ Highways that descend steeply areoften provided with escape ramps thatenable truck drivers whose brakes fail tobring their rigs to a safe stop. These rampsprovide a perfect illustration of the con-servation of energy. From a physics pointof view, the driver’s problem is to get ridof an enormous amount of kinetic energyin the safest possible way. The ramps runuphill, so some of the kinetic energy issimply converted back into gravitationalpotential energy (just as in a rollercoaster). In addition, the ramps are typi-cally surfaced with sand or gravel, allow-ing much of the initial kinetic energy to bedissipated by friction into other forms ofenergy, such as sound and heat.

hA D

C

B

x

y

FIGURE 8–10 A ball rolling on africtionless trackThe ball starts at A, where withzero speed. Its greatest speed occurs at B.At D, where again, its speedreturns to zero.

y = h

y = h,

▲

8–5 Potential Energy Curves and EquipotentialsFigure 8–10 shows a metal ball rolling on a roller coaster-like track. Initially theball is at rest at point A. Since the height at A is the ball’s initial mechanicalenergy is If friction and other nonconservative forces can be ignored,the ball’s mechanical energy remains fixed at throughout its motion. Thus,

As the ball moves, its potential energy falls and rises in the same way as thetrack. After all, the gravitational potential energy, is directly proportionalU = mgy,

E = U + K = E0

E0

E0 = mgh.y = h,

WALKMC08_0131536311.QXD 12/8/05 17:48 Page 227

The ball speeds up as it drops lower.

Energy Conservation and Energy Diagrams

This can be represented with a potential energy curve of the sameshape!

228 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

to the height of the track, y. In a sense, then, the track itself represents a graph of thecorresponding potential energy.

This is shown explicitly in Figure 8–11, where we plot energy on the verticalaxis and x on the horizontal axis. The potential energy U looks just like the track inFigure 8–10. In addition, we plot a horizontal line at the value indicating theconstant energy of the ball. Since the potential energy plus the kinetic energy mustalways add up to it follows that K is the amount of energy from the potential en-ergy curve up to the horizontal line at This is also shown in Figure 8–11.

Examining an energy plot like Figure 8–11 gives a great deal of informationabout the motion of an object. For example, at point B the potential energy has itslowest value, and thus the kinetic energy is greatest there. At point C the potentialenergy has increased, indicating a corresponding decrease in kinetic energy. As theball continues to the right, the potential energy increases until, at point D, it is againequal to the total energy, At this point the kinetic energy is zero, and the ballcomes to rest momentarily. It then “turns around” and begins to move to the left,eventually returning to point A where it again stops, changes direction, and beginsa new cycle. Points A and D, then, are referred to as turning points of the motion.

Turning points are also seen in the motion of a mass on a spring, as indicatedin Figure 8–12. Figure 8–12 (a) shows a mass pulled to the position andreleased from rest; Figure 8–12 (b) shows the potential energy of the system,

Starting the system this way gives it an initial energy shownE0 = 12 kA2,U = 1

2 kx2.

x = A,

E0.

E0.E0,

E0,

E0 = mghA D

C

B

xU

K

U

K

U

K

Ener

gy

FIGURE 8–11 Gravitational potentialenergy versus position for the trackshown in Figure 8–10The shape of the potential energy curveis exactly the same as the shape of thetrack. In this case, the total mechanicalenergy is fixed at its initial value,

Because the heightof the curve is U, by definition, it followsthat K is the distance from the curve upto the dashed line at Note thatK is largest at B. In addition, K vanishesat A and D, which are turning points ofthe motion.

E0 = mgh.

E0 = U + K = mgh.

▲

FIGURE 8–12 A mass on a spring(a) A spring is stretched by an amount A,giving it a potential energy of(b) The potential energy curve,for the spring in (a). Because the massstarts at rest, its initial mechanical energyis The mass oscillates between

and x = - A.x = AE0 = 1

2 kA2.

U = 12 kx2,

U = 12 kA2.

▲

x0

0

total mechanical energy

turning pointturning point

x = Ax = –A

x = Ax = –A

E0 = kA21–2

U

(a)

(b)

x

mm

WALKMC08_0131536311.QXD 12/8/05 17:48 Page 228

The lower points correspond to less potential energy and morekinetic energy.

If the initial PE is E0, then K + U = E0 at all points.

Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are stable equilibria?


force of friction, and (ii) the force generated by the car’s engine.The work done by friction is the work done bythe engine is Find the change in the car’s kineticenergy from the bottom of the hill to the top of the hill.

37. •• IP An 81.0-kg in-line skater does of nonconserva-tive work by pushing against the ground with his skates. In ad-dition, friction does of nonconservative work on theskater. The skater’s initial and final speeds are 2.50 m/s and1.22 m/s, respectively. (a) Has the skater gone uphill, downhill,or remained at the same level? Explain. (b) Calculate thechange in height of the skater.

38. •• In Example 8–10, suppose the two masses start from restand are moving with a speed of 2.05 m/s just before hits thefloor. (a) If the coefficient of kinetic friction is whatis the distance of travel, d, for the masses? (b) How much con-servative work was done on this system? (c) How much non-conservative work was done on this system? (d) Verify the threework relations given in Equation 8–10.

39. •• IP A 15,800-kg truck is moving at 12.0 m/s when it startsdown a 6.00° incline in the Canadian Rockies. At the start of thedescent the driver notices that the altitude is 1630 m. When shereaches an altitude of 1440 m, her speed is 29.0 m/s. Find thechange in (a) the gravitational potential energy of the systemand (b) the truck’s kinetic energy. (c) Is the total mechanical en-ergy of the system conserved? Explain.

40. ••• A 1.80-kg block slides on a rough horizontal surface. Theblock hits a spring with a speed of 2.00 m/s and compressesit a distance of 11.0 cm before coming to rest. If the coefficientof kinetic friction between the block and the surface is

what is the force constant of the spring?

Section 8–5 Potential Energy Curves and Equipotentials41. • Figure 8–24 shows a potential energy curve as a function of

x. In qualitative terms, describe the subsequent motion of anobject that starts at rest at point A.

mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;

system as a function of the angle the ropes make with the verti-cal, assuming the potential energy is zero when the ropes arevertical. Consider angles up to 90° on either side of the vertical.

45. •• Find the turning-point angles in the previous problem if thechild has a speed of 0.89 m/s when the ropes are vertical. Indi-cate the turning points on a plot of the system’s potential energy.

46. •• The potential energy of a particle moving along the x axis isshown in Figure 8–24. When the particle is at it has3.6 J of kinetic energy. Give approximate answers to the follow-ing questions. (a) What is the total mechanical energy of thesystem? (b) What is the smallest value of x the particle canreach? (c) What is the largest value of x the particle can reach?

47. •• A block of mass is connected to a spring offorce constant on a smooth, horizontal surface.(a) Plot the potential energy of the spring from to (b) Determine the turning points of the block ifits speed at is 1.3 m/s.

48. •• A ball of mass is thrown straight upward withan initial speed of 8.9 m/s. (a) Plot the gravitational potentialenergy of the block from its launch height, to the height

Let correspond to (b) Determine theturning point (maximum height) of this mass.

49. ••• Two blocks, each of mass m, are connected on a frictionlesshorizontal table by a spring of force constant k and equilibriumlength L. Find the maximum and minimum separation betweenthe two blocks in terms of their maximum speed, relativeto the table. (The two blocks always move in opposite direc-tions as they oscillate back and forth about a fixed position.)

General Problems50. •• IP A sled slides without friction down a small, ice-covered

hill. If the sled starts from rest at the top of the hill, its speed atthe bottom is 7.50 m/s. (a) On a second run, the sled starts witha speed of 1.50 m/s at the top. When it reaches the bottom of thehill, is its speed 9.00 m/s, more than 9.00 m/s, or less than9.00 m/s? Explain. (b) Find the speed of the sled at the bottomof the hill after the second run.

51. •• In the previous problem, what is the height of the hill?

52. •• A 61-kg skier encounters a dip in the snow’s surface thathas a circular cross section with a radius of curvature of 12 m. Ifthe skier’s speed at point A in Figure 8–25 is 8.0 m/s, what isthe normal force exerted by the snow on the skier at point B?Ignore frictional forces.

vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m

▲ FIGURE 8–24 Problems 41, 42, 43, and 46

42. • An object moves along the x axis, subject to the potential en-ergy shown in Figure 8–24. The object has a mass of 1.1 kg andstarts at rest at point A. (a) What is the object’s speed at point B?(b) At point C? (c) At point D? (d) What are the turning pointsfor this object?

43. • A 1.44-kg object moves along the x axis, subject to the poten-tial energy shown in Figure 8–24. If the object’s speed at point Cis 1.25 m/s, what are the approximate locations of its turningpoints?

44. • A 23-kg child swings back and forth on a swing suspended by2.0-m-long ropes. Plot the gravitational potential energy of this

r = 12 m

1.75 m A

B

▲ FIGURE 8–25 Problem 52

53. •• The spring in a clothespin is compressed 0.40 cm, storing en-ergy in the form of spring potential energy. If the spring is com-pressed twice as far, the spring potential energy increases by0.0046 J. What is the force constant, k, for this spring?

54. •• In a circus act, a 66-kg trapeze artist starts from rest with the4.5-m trapeze rope horizontal. What is the tension in the ropewhen it is vertical?

WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

(A) A and E(B) B and D(C) C(D) B, C, and D

Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are stable equilibria?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

(A) A and E(B) B and D ←(C) C(D) B, C, and D

Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are unstable equilibria?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236


Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are unstable equilibria?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

(A) A and E(B) B and D(C) C ←(D) B, C, and D

Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are turning points?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236


Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest at pointA. Which points are turning points?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

(A) A and E ←(B) B and D(C) C(D) B, C, and D

Energy Conservation Diagrams ExamplesAn object moves along the x axis, subject to the potential energyshown. The object has a mass of 1.1 kg and starts at rest atpoint A.(a) What is the object’s speed at point B? (b) At point C?










mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

Hypoth: greater at B than C (a) v = 3.8 m/s ; (b) v = 2.7 m/s











mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

Hypoth: greater at B than C

(a) v = 3.8 m/s ; (b) v = 2.7 m/s











mk = 0.560,

mk = 0.350,m2

- 715 J

+ 3420 J

+ 6.44 * 105 J.- 3.11 * 105 J;












vmax,

y = 0.U = 0y = 5.0 m.y = 0,

m = 0.75 kg

x = 0x = 5.00 cm.

x = - 5.00 cmk = 775 N/m

m = 0.95 kg

x = 1.0 m

x

U

A

B

CD

E

2.0 J

5.0 J6.0 J

10.0 J

5.0 m4.0 m3.0 m2.0 m1.0 m





r = 12 m

1.75 m A

B




WALKMC08_0131536311.QXD 12/8/05 17:48 Page 236

Hypoth: greater at B than C (a) v = 3.8 m/s ; (b) v = 2.7 m/s

How to Solve Energy Conservation Problems

1 Draw (a) diagram(s). Free body diagrams or full pictures, asneeded.

2 Make a hypothesis or estimate of what the answer will be.

3 Identify the system. State what it is. Is it isolated?

4 Identify the initial point / configuration of the system.

5 Identify the final point / configuration of the system.

6 Write the energy conservation equation.

7 Fill in the expressions as needed.

8 Solve.

9 Analyze answer: reasonable value?, check units, etc.

Energy Conservation Example


and, (ii) whenever possible, describe a natural pro-cess in which the energy transfer or transformation occurs. Give details to defend your choices, such as identifying the system and identifying other output energy if the device or natural process has limited efficiency. (a) Chemical potential energy transforms into internal energy. (b) Energy transferred by elec-trical transmission becomes gravitational potential energy. (c) Elastic potential energy transfers out of a system by heat. (d) Energy transferred by mechani-cal waves does work on a system. (e) Energy carried by electromagnetic waves becomes kinetic energy in a system.

9. A block is connected to a spring that is suspended from the ceiling. Assuming air resistance is ignored, describe the energy transformations that occur within the system consisting of the block, the Earth, and the spring when the block is set into vertical motion.

10. In Chapter 7, the work–kinetic energy theorem, W 5 DK, was introduced. This equation states that work done on a system appears as a change in kinetic energy. It is a special-case equation, valid if there are no changes in any other type of energy such as potential or internal. Give two or three examples in which work is done on a system but the change in energy of the system is not a change in kinetic energy.

tip of the demonstrator’s nose as shown in Figure CQ8.5. The dem-onstrator remains stationary. (a) Ex- plain why the ball does not strike her on its return swing. (b) Would this demonstrator be safe if the ball were given a push from its starting position at her nose?

6. Can a force of static friction do work? If not, why not? If so, give an example.

7. In the general conservation of energy equation, state which terms predominate in describing each of the following devices and processes. For a process going on continu-ously, you may consider what happens in a 10-s time interval. State which terms in the equation represent original and final forms of energy, which would be inputs, and which outputs. (a) a slingshot firing a peb-ble (b) a fire burning (c) a portable radio operating (d) a car braking to a stop (e) the surface of the Sun shining visibly (f) a person jumping up onto a chair

8. Consider the energy transfers and transformations listed below in parts (a) through (e). For each part, (i) describe human-made devices designed to pro-duce each of the energy transfers or transformations

Section 8.1 Analysis Model: Nonisolated System (Energy) 1. For each of the following systems and time intervals,

write the appropriate version of Equation 8.2, the conservation of energy equation. (a) the heating coils in your toaster during the first five seconds after you turn the toaster on (b) your automobile from just before you fill it with gasoline until you pull away from the gas station at speed v (c) your body while you sit quietly and eat a peanut butter and jelly sand-wich for lunch (d) your home during five minutes of a sunny afternoon while the temperature in the home remains fixed

2. A ball of mass m falls from a height h to the floor. (a) Write the appropriate version of Equation 8.2 for the system of the ball and the Earth and use it to cal-culate the speed of the ball just before it strikes the Earth. (b) Write the appropriate version of Equation 8.2 for the system of the ball and use it to calculate the speed of the ball just before it strikes the Earth.

S

S

Section 8.2 Analysis Model: Isolated System (Energy) 3. A block of mass 0.250 kg is placed on top of a light, ver-

tical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

4. A 20.0-kg cannonball is fired from a cannon with muz-zle speed of 1 000 m/s at an angle of 37.08 with the hor-izontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechani-cal energy of the ball–Earth sys-tem at the maximum height for each ball. Let y 5 0 at the cannon.

5. Review. A bead slides without fric-tion around a loop-the-loop (Fig. P8.5). The bead is released from rest at a height h 5 3.50R. (a) What

W

W

AMTM

Figure CQ8.5

Problems

The problems found in this

chapter may be assigned online in Enhanced WebAssign

1. straightforward; 2. intermediate; 3. challenging

1. full solution available in the Student Solutions Manual/Study Guide

AMT Analysis Model tutorial available in Enhanced WebAssign

GP Guided Problem

M Master It tutorial available in Enhanced WebAssign

W Watch It video solution available in Enhanced WebAssign

BIO

Q/C

S

hR

!

Figure P8.5

Wapp → Us =1

2kx2 → K.E. & Grav P.E.→ Ug = mgh

System: block + spring + Earth.

Initial point, i©: release point (max compression of spring),choose y = 0, U = 0 at this point

Final point, f©: point of max height of block

System is isolated.

Energy Conservation Example


and, (ii) whenever possible, describe a natural pro-cess in which the energy transfer or transformation occurs. Give details to defend your choices, such as identifying the system and identifying other output energy if the device or natural process has limited efficiency. (a) Chemical potential energy transforms into internal energy. (b) Energy transferred by elec-trical transmission becomes gravitational potential energy. (c) Elastic potential energy transfers out of a system by heat. (d) Energy transferred by mechani-cal waves does work on a system. (e) Energy carried by electromagnetic waves becomes kinetic energy in a system.

9. A block is connected to a spring that is suspended from the ceiling. Assuming air resistance is ignored, describe the energy transformations that occur within the system consisting of the block, the Earth, and the spring when the block is set into vertical motion.

10. In Chapter 7, the work–kinetic energy theorem, W 5 DK, was introduced. This equation states that work done on a system appears as a change in kinetic energy. It is a special-case equation, valid if there are no changes in any other type of energy such as potential or internal. Give two or three examples in which work is done on a system but the change in energy of the system is not a change in kinetic energy.

tip of the demonstrator’s nose as shown in Figure CQ8.5. The dem-onstrator remains stationary. (a) Ex- plain why the ball does not strike her on its return swing. (b) Would this demonstrator be safe if the ball were given a push from its starting position at her nose?

6. Can a force of static friction do work? If not, why not? If so, give an example.

7. In the general conservation of energy equation, state which terms predominate in describing each of the following devices and processes. For a process going on continu-ously, you may consider what happens in a 10-s time interval. State which terms in the equation represent original and final forms of energy, which would be inputs, and which outputs. (a) a slingshot firing a peb-ble (b) a fire burning (c) a portable radio operating (d) a car braking to a stop (e) the surface of the Sun shining visibly (f) a person jumping up onto a chair

8. Consider the energy transfers and transformations listed below in parts (a) through (e). For each part, (i) describe human-made devices designed to pro-duce each of the energy transfers or transformations

Section 8.1 Analysis Model: Nonisolated System (Energy) 1. For each of the following systems and time intervals,

write the appropriate version of Equation 8.2, the conservation of energy equation. (a) the heating coils in your toaster during the first five seconds after you turn the toaster on (b) your automobile from just before you fill it with gasoline until you pull away from the gas station at speed v (c) your body while you sit quietly and eat a peanut butter and jelly sand-wich for lunch (d) your home during five minutes of a sunny afternoon while the temperature in the home remains fixed

2. A ball of mass m falls from a height h to the floor. (a) Write the appropriate version of Equation 8.2 for the system of the ball and the Earth and use it to cal-culate the speed of the ball just before it strikes the Earth. (b) Write the appropriate version of Equation 8.2 for the system of the ball and use it to calculate the speed of the ball just before it strikes the Earth.

S

S

Section 8.2 Analysis Model: Isolated System (Energy) 3. A block of mass 0.250 kg is placed on top of a light, ver-

tical spring of force constant 5 000 N/m and pushed downward so that the spring is compressed by 0.100 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

4. A 20.0-kg cannonball is fired from a cannon with muz-zle speed of 1 000 m/s at an angle of 37.08 with the hor-izontal. A second ball is fired at an angle of 90.08. Use the isolated system model to find (a) the maximum height reached by each ball and (b) the total mechani-cal energy of the ball–Earth sys-tem at the maximum height for each ball. Let y 5 0 at the cannon.

5. Review. A bead slides without fric-tion around a loop-the-loop (Fig. P8.5). The bead is released from rest at a height h 5 3.50R. (a) What

W

W

AMTM

Figure CQ8.5

Problems

The problems found in this

chapter may be assigned online in Enhanced WebAssign

1. straightforward; 2. intermediate; 3. challenging

1. full solution available in the Student Solutions Manual/Study Guide

AMT Analysis Model tutorial available in Enhanced WebAssign

GP Guided Problem

M Master It tutorial available in Enhanced WebAssign

W Watch It video solution available in Enhanced WebAssign

BIO

Q/C

S

hR

!

Figure P8.5

∆K + ∆U = 0

(��>0

Kf −��0

Ki ) + (��*0

Us,f − Us,i ) + (Ug ,f −��*0

Ug ,i ) = 0

Ug ,f = Us,i

mgh =1

2kx2

h =kx2

2mgh = 10.2 m

1Problem from Serway & Jewett, 9th ed, page 236.

Energy Conservation: Example 8-10A block of mass m1 = 2.40 kg is connected to a second block ofmass m2 = 1.80 kg. When the blocks are released from rest, theymove through a distance d = 0.500 m, at which point m2 hits thefloor. Given that the coefficient of kinetic friction between m1 andthe horizontal surface is µk = 0.450, find the speed of the blocksjust before m2 lands.

EXAMPLE 8–10 Landing with a ThudA block of mass is connected to a second block of mass as shown here. When the blocks are releasedfrom rest, they move through a distance at which point hits the floor. Given that the coefficient of kinetic fric-tion between and the horizontal surface is find the speed of the blocks just before lands.

Picture the ProblemWe choose to be at floor level; therefore, the gravita-tional potential energy of is zero when it lands. The poten-tial energy of doesn’t change during this process; it isalways Thus, it isn’t necessary to know the value of h.Note that we label the beginning and ending points with i andf, respectively.

StrategySince a nonconservative force (friction) is doing work in thissystem, we use Thus, we must calculatenot only the mechanical energies, and but also the non-conservative work, Note that can be written in terms ofthe unknown speed of the blocks just before lands. There-fore, we can set equal to and solve for the final speed.

Solution

1. Evaluate and Be sure to include contributions from both masses:

2. Next, evaluate and Note that depends on the unknown speed, v:

3. Calculate the nonconservative work, Recall that the force of friction is and that it points opposite to the displacement of distance d:

4. Set equal to Notice that cancels because it occurs in both and

5. Solve for v:

6. Substitute numerical values:

InsightNote that step 4 can be rearranged as follows: Translating this to words, we can say that thefinal kinetic energy of the blocks is equal to the initial gravitational potential energy of minus the energy dissipated by friction.

Practice ProblemFind the coefficient of kinetic friction if the final speed of the blocks is 0.950 m/s. [Answer: ]

Some related homework problems: Problem 35, Problem 36

mk = 0.589

m2,

12 m1v2 + 1

2 m2v2 = m2gd - mkm1gd.

= 1.30 m/s

v = A2[1.80 kg - 10.450212.40 kg2]19.81 m/s2210.500 m21.80 kg + 2.40 kg

v = A21m2 - mkm12gd

m1 + m2

-mkm1gd = 12 m1v2 + 1

2 m2v2 - m2gdEf.Ei

Wnc = Ef - Eim1gh¢E = Ef - Ei.Wnc

fk = mkN = mkm1g,Wnc = -fkd = -mkm1gdWnc.

Ef = Uf + Kf = m1gh + 12 m1v2 + 1

2 m2v2

Kf = 12 m1v2 + 1

2 m2v2

Uf = m1gh + 0EfEf.Uf, Kf,

Ei = Ui + Ki = m1gh + m2gd

Ki = 12 m1 # 02 + 1

2 m2 # 02 = 0

Ui = m1gh + m2gdEi.Ui, Ki,

¢EWnc

m2

EfWnc.Ef,Ei

Wnc = ¢E = Ef - Ei.

m1gh.m1

m2

y = 0

m2mk = 0.450,m1

m2d = 0.500 m,m2 = 1.80 kg,m1 = 2.40 kg


m2

i f

d

d

0

h

f

i

y

m1

Finally, we present an Active Example for the common situation of a system inwhich two different nonconservative forces do work.

WALKMC08_0131536311.QXD 12/8/05 17:48 Page 226

Example 8-10EXAMPLE 8–10 Landing with a ThudA block of mass is connected to a second block of mass as shown here. When the blocks are releasedfrom rest, they move through a distance at which point hits the floor. Given that the coefficient of kinetic fric-tion between and the horizontal surface is find the speed of the blocks just before lands.

Picture the ProblemWe choose to be at floor level; therefore, the gravita-tional potential energy of is zero when it lands. The poten-tial energy of doesn’t change during this process; it isalways Thus, it isn’t necessary to know the value of h.Note that we label the beginning and ending points with i andf, respectively.

StrategySince a nonconservative force (friction) is doing work in thissystem, we use Thus, we must calculatenot only the mechanical energies, and but also the non-conservative work, Note that can be written in terms ofthe unknown speed of the blocks just before lands. There-fore, we can set equal to and solve for the final speed.

Solution

1. Evaluate and Be sure to include contributions from both masses:

2. Next, evaluate and Note that depends on the unknown speed, v:

3. Calculate the nonconservative work, Recall that the force of friction is and that it points opposite to the displacement of distance d:

4. Set equal to Notice that cancels because it occurs in both and

5. Solve for v:

6. Substitute numerical values:

InsightNote that step 4 can be rearranged as follows: Translating this to words, we can say that thefinal kinetic energy of the blocks is equal to the initial gravitational potential energy of minus the energy dissipated by friction.

Practice ProblemFind the coefficient of kinetic friction if the final speed of the blocks is 0.950 m/s. [Answer: ]

Some related homework problems: Problem 35, Problem 36

mk = 0.589

m2,

12 m1v2 + 1

2 m2v2 = m2gd - mkm1gd.

= 1.30 m/s

v = A2[1.80 kg - 10.450212.40 kg2]19.81 m/s2210.500 m21.80 kg + 2.40 kg

v = A21m2 - mkm12gd

m1 + m2

-mkm1gd = 12 m1v2 + 1

2 m2v2 - m2gdEf.Ei

Wnc = Ef - Eim1gh¢E = Ef - Ei.Wnc

fk = mkN = mkm1g,Wnc = -fkd = -mkm1gdWnc.

Ef = Uf + Kf = m1gh + 12 m1v2 + 1

2 m2v2

Kf = 12 m1v2 + 1

2 m2v2

Uf = m1gh + 0EfEf.Uf, Kf,

Ei = Ui + Ki = m1gh + m2gd

Ki = 12 m1 # 02 + 1

2 m2 # 02 = 0

Ui = m1gh + m2gdEi.Ui, Ki,

¢EWnc

m2

EfWnc.Ef,Ei

Wnc = ¢E = Ef - Ei.

m1gh.m1

m2

y = 0

m2mk = 0.450,m1

m2d = 0.500 m,m2 = 1.80 kg,m1 = 2.40 kg


m2

i f

d

d

0

h

f

i

y

m1

Finally, we present an Active Example for the common situation of a system inwhich two different nonconservative forces do work.

WALKMC08_0131536311.QXD 12/8/05 17:48 Page 226

System: Masses m1 and m2, modeled as point particles, and theEarth.

Wext = ∆K + ∆U

Here:Wext = Wnc = −fkd∆K is the change in K.E. of both masses∆U is the change in Grav. P.E. of the masses (only m2’s changes)

Example 8-10

Points i© and f© are as labelled in the diagram.

Wext = ∆K + ∆U

Wext = (K1,f + K2,f −��*0

K1,i −��*0

K2,i ) + (��>0

Uf − Ui )

−fkd = (1

2(m1 +m2)v

2 − 0) + (0 −m2gd)

v =

√2(m2 − µkm1)gd

m1 +m2

= 1.30 m/s

Summary

• energy conservation practice

Final Exam Thursday, Mar 29, 9:15-11:15am, S35 (here).

HomeworkWalker Physics:

• PREV: Ch 8, onward from page 243. Questions: 11, 13;Problems: 21, 23, 37, 41, 47, 55, 57, 59, 87, 95

• NEW: Ch 8, Problem: 97 (Only 1 new problem. Notice in thisproblem, the ramp is curved, so the acceleration of the blockis not constant, which means you cannot use kinematicsequations. You must use conservation of energy.)

introduction to mechanics energy...

Documents